112.path-sum 路径总和

带明显的回溯的版本

#include <vector>
using std::vector;
class Solution {
  private:
    vector<int> res;
    int sum = 0;

  public:
    void cnt_sum(TreeNode *root) {
        if (root->left == nullptr && root->right == nullptr) {
            sum += root->val;
            res.push_back(sum);
            return;
        }
        sum += root->val;
        if (root->left != nullptr) {
            cnt_sum(root->left);
            sum -= root->left->val;//回溯
        }
        if (root->right != nullptr) {
            cnt_sum(root->right);
            sum -= root->right->val;//回溯
        }
        return;
    }
    bool hasPathSum(TreeNode *root, int targetSum) {
        if (root == nullptr)
            return false;
        cnt_sum(root);
        for (int i : res) {
            if (i == targetSum)
                return true;
        }
        return false;
    }
};

回溯被隐藏起来的版本

#include <vector>
using std::vector;
class Solution {
  private:
    vector<int> res;
    //int sum = 0;

  public:
    void cnt_sum(TreeNode *root, int sum) {
        if (root->left == nullptr && root->right == nullptr) {
            sum += root->val;
            res.push_back(sum);
            return;
        }
        sum += root->val;
        if (root->left != nullptr) {
            cnt_sum(root->left, sum); //注意这里的sum作为实参，其值被改变并不会改变实际上sum的值
            //sum -= root->left->val;
        }
        if (root->right != nullptr) {
            cnt_sum(root->right, sum);
            //sum -= root->right->val;
        }
        return;
    }
    bool hasPathSum(TreeNode *root, int targetSum) {
        if (root == nullptr)
            return false;
        cnt_sum(root, 0);
        for (int i = 0; i < res.size(); i++)
            if (res[i] == targetSum)
                return true;
        return false;
    }
};